Question 1180006
A soccer ball was hit and moved upward at a 50.0° angle with an initial velocity of 40.0 m/s.
a. Find the length of time of flight.
b. What is the horizontal distance reached by the soccer ball?
c. What is the maximum vertical distance reached by the soccer ball?
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<pre>
Having given the vector v as v = (40 m/s, 50°), calculate first its horizontal and verical components


    {{{V[hor]}}}  = 40*cos(50°) = 40*0.643 = 25.725 m/s;

    {{{V[vert]}}} = 40*sin(50°) = 40*0.765 = 30.630 m/s.


(a)  the length of time flight {{{t[flight]}}} = {{{2*(V[vert]/g)}}} = {{{2*(30.630/9.81)}}} = 6.245 seconds.


     Here g = 9.81 m/s^2 is the gravity acceleration.



(b)  the horizontal distance  {{{d[hor]}}} = {{{V[hor]*t[flight]}}} = 25.725*6.245 = 160.65 meters.



(c)  the maximum vertical distance  height = {{{V[vert]^2/(2g)}}} = {{{30.630^2/(2*9.81)}}} = 47.82 meters.
</pre>

Solved.


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